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I can not get the correct answer.

$$\int \frac {dx}{\sqrt {x^2 + 16}}$$

$x = 4 \tan \theta$, $dx = 4\sec^2 \theta$

$$\int \frac {dx}{\sqrt {16 \sec^2 \theta}}$$

$$\int \frac {4 \sec^ 2 \theta}{\sqrt {16 \sec^2 \theta}}$$

$$\int \frac {4 \sec^ 2 \theta}{4 \sec \theta}$$

$$\int \sec \theta$$

$$\ln| \sec \theta + \tan \theta|$$

Then I solve for $\theta$:

$x = 4 \tan \theta$

$x/4 = \tan \theta$

$\arctan (\frac{x}{4}) = \theta$

$$\ln| \sec (\arctan (\tfrac{x}{4})) + \tan (\arctan (\tfrac{x}{4}))|$$

$$\ln| \sec (\arctan (\tfrac{x}{4})) + \tfrac{x}{4}))| + c$$

This is wrong and I do not know why.

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2  
It looks good. All that's left to do is simplify $\sec(\arctan(x/4))$. –  David Mitra Jun 4 '12 at 15:43
    
Looks pretty good to me. But $\sec^2\theta=1+\tan^2\theta=1+(x/4)^2$, so $\sec\theta=\sqrt{1+(x/4)^2}=\frac{1}{4}\sqrt{16+x^2}$. (I am being sloppy and not worrying about signs. Should!) Can simplify further, using logarithm laws, and changing the arbitrary constant. –  André Nicolas Jun 4 '12 at 15:43
    
Yes, you just have it in a slightly different form than would normally be used. Also, there are some slight mistakes in your notation throughout the process. –  process91 Jun 4 '12 at 15:45

1 Answer 1

up vote 8 down vote accepted

What you have done is correct! Note that whenever you have inverse trigonometric expressions you can express your answer in more than one way! Your answer can be expressed in a different way (without the trigonometric and inverse trigonometric functions) as shown below.

We will prove that $$\sec \left( \arctan \left( \dfrac{x}4 \right) \right) = \sqrt{1 + \left(\dfrac{x}{4} \right)^2}$$

Hence, your answer $$\ln \left \lvert \dfrac{x}4 + \sec \left(\arctan \left( \dfrac{x} 4\right) \right) \right \rvert + c$$ can be rewritten as $$\ln \left \lvert \dfrac{x}4 + \sqrt{1+ \left(\dfrac{x}{4} \right)^2} \right \rvert + c$$ Note that $$\theta = \arctan\left( \dfrac{x}4 \right) \implies \tan( \theta) = \dfrac{x}4 \implies \tan^2(\theta) = \dfrac{x^2}{16} \implies 1 + \tan^2(\theta) = 1+\dfrac{x^2}{16}$$ Hence, we get that $$\sec^2(\theta) = 1+ \left(\dfrac{x}{4} \right)^2 \implies \sec (\theta) = \sqrt{1+ \left(\dfrac{x}{4} \right)^2} \implies \sec \left(\arctan\left( \dfrac{x}4 \right) \right) = \sqrt{1+ \left(\dfrac{x}{4} \right)^2}$$ Hence, you can rewrite your answer as $$\ln \left \lvert \dfrac{x}4 + \sqrt{1+ \left(\dfrac{x}{4} \right)^2} \right \rvert + c$$

Also, you have been a bit sloppy with some notations in your argument.

For instance, when you substitute $x = 4 \tan (\theta)$, $$\dfrac{dx}{\sqrt{x^2+16}} \text{ should immediately become }\dfrac{4 \sec^2(\theta)}{\sqrt{16 \sec^2(\theta)}} d \theta$$

Also, you need to carry the $d \theta$ throughout the answer under the integral.

Writing just $\displaystyle \int\sec(\theta)$ or $\displaystyle \int\dfrac{4 \sec^2(\theta)}{\sqrt{16 \sec^2(\theta)}}$ without the $d \theta$ is notationally incorrect.

Anyway, I am happy that you are slowly getting a hang of these!

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